Question

$X{X}^{\prime }$ is a conducting rod which can slide, freely on U shaped frame with resistance $R$. A constant magnetic field $B$ acts perpendicular to the frame. If a constant force is applied on a string connected to the rod as shown in figure, such that the rod moves with a constant velocity of $V_0.$ If the direction of velocity and magnetic field are reversed without changing their magnitudes and if the constant force is applied to move to rod $X{X}^{\prime }$ with the same constant speed, then

• A. The magnetic force on $X{X}^{\prime }$ is $\frac{B^2{l}^2{V}_0}{R}\left(\hat{i}\right)$ before reversing velocity and magnetic fields.
• B. The magnetic force on $X{X}^{\prime }$ is $\frac{B^2{l}^2{V}_0}{R}\left(\widehat{+j}\right)$before reversing velocity and magnetic fields.
• C. The magnetic force on $X{X}^{\prime }$ is $\frac{B^2{l}^2{V}_0}{R}\left(\widehat{+i}\right)$ after reversing velocity and magnetic fields.
• D. The magnetic force on $X{X}^{\prime }$ is $\frac{B^2{l}^2{V}_0}{R}\left(\widehat{-j}\right)$ after reversing velocity and magnetic fields.

Answer 1

Answer: (B), (D)

The magnetic force on $X{X}^{\prime }$ is $\frac{B^2{l}^2{V}_0}{R}\left(\widehat{-j}\right)$ after reversing velocity and magnetic fields.

As we know that induced emf in a conductor moving through magnetic fields given by $e=B{V}_{0}l$
Case 1




$\therefore$ induced current is
$I=\frac{e}{R}=\frac{B{V}_{0}l}{R}$
So the magnetic force on $X{X}^{\prime }$
$F=ilB$
$F=\frac{B{V}_{0}l}{R}.lB$
$\Rightarrow F=\frac{B^2{V}_0{l}^2}{R}$
By left hand rule,
$\overrightarrow{F}=\frac{B^2{V}_{0}l}{R}\left(\hat{j}\right)$
Case -II



The direction of induced current by the fleming's right rule which is downward in direction.
$i=\frac{Bl{V}_0}{R}$
Therefore the magnetic force is given by
$\overrightarrow{F}=Bil(-\hat{j})=\frac{B^2{l}^2{V}_0}{R}(-\hat{j})$ {Direction of force is obtained by left hand rule }